Angular Velocity
Announcements:
•  Midterm on Thursday at 7:30pm! Old exams
available on website. Chapters 6–9 are covered.
Go to same room as last time.
–  You are allowed one calculator and one doublesided sheet of paper with hand written notes
–  14 Multiple choice plus two long answer questions
–  Test time from 7:30pm – 9:00pm. If you have a
conflict, need extra time, etc., then contact
Professor Daniel Dessau.
Web page: http://www.colorado.edu/physics/phys1110/phys1110_sp12/
Clicker Score Update
•  Clickers are posted on D2L up to March 9.
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These clickers are being
used in recently
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recorded responses, but
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they are not registered to
any students in the class.
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Please check your
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clicker score on D2L!
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Exam Rooms
Clicker question 1
Set frequency to BA
Steve (m=50. kg) is skating east with velocity
v=1.0 m/s . Scott (m=70. kg) is skating north
with velocity v=1.0 m/s . They happen to collide
at the center of the rink, and hold tight (i.e. they
“stick together”, oomph!) What is the final speed
of the pair, after the collision?
A) 1.4 m/s
B) 1.0 m/s C) 0.5 m/s
D) 0.0 m/s
E) 0.72 m/s
Clicker question 1
Set frequency to BA
Steve (m=50. kg) is skating east with velocity
v=1.0 m/s . Scott (m=70. kg) is skating north
with velocity v=1.0 m/s . They happen to collide
at the center of the rink, and hold tight (i.e. they
“stick together”, oomph!) What is the final speed
of the pair, after the collision?
A) 1.4 m/s
B) 1.0 m/s C) 0.5 m/s
D) 0.0 m/s
E) 0.72 m/s

ˆ
ˆ
(50kg)(1m /s)i + (70kg)(1m /s) j = (120kg)v f

5 ˆ 7 ˆ
vf = i +
j
12 12

2
2
v = (5 /12) + (7 /12) ≈ .72m /s
Clicker question 2
Set frequency to BA
Four identical square tiles are glued together
like a “J”, as shown. Each tile has edge
length “a” and mass “m”. The origin is at
the bottom right, as shown. What is the x
component of the center of mass of the
system of tiles?
A) -(1/2) a
B) -(3/4) a
C) - a
D) -(3/2) a
E) -3 a
Clicker question 2
Set frequency to BA
Four identical square tiles are glued together
like a “J”, as shown. Each tile has edge
length “a” and mass “m”. The origin is at
the bottom right, as shown. What is the x
component of the center of mass of the
system of tiles?
A) -(1/2) a
B) -(3/4) a
C) - a
D) -(3/2) a
E) -3 a
x(cm) = (m1 x1 + m2 x + m3 x3 + m4 x4 ) / (mtotal)
Numbering the squares 1, 2, and 3 walking down, and #4 is the one on
the left, gives x(cm) = m*(-a/2)+m*(-a/2)+m*(-a/2)+m*(-3/2a)/ (4m) =
-3a/4.
€
Using concept of negative mass
A square of side 2R has a circular hole of
radius R/2 removed. Relative to the center of
the square the center of the hole is located at
(R/2,R/2). Locate the center of mass with
respect to the center of the square.
CoM =
∑ mi x i
∑m
i
msq x sq − mcir x cir
CoM =
msq − mcir
Treat the hole as negative mass:
x(cm) = [4R2 (0) – π (R/2)2 (R/2)]/[4R2-π(R/2)2]
πR
π
€
−
/[4 − ] = −.393R /3.21 = −.122R
8
4
y(cm)=-.122R (same as x(cm)
R/2
2R
Angular kinematics
In chapters 2 and 3 we dealt with kinematics which
involved displacement, velocity, and acceleration.
Angular kinematics is the same thing but for objects
which are rotating (rather than translating).
For something to rotate, it must have an axis
about which it rotates like the axle for a wheel.
Only sensible place for the origin is along the axis.
Use polar coordinates (r,θ) instead
of Cartesian coordinates (x,y).
Axis is in the z-direction.
r
θ
Angular velocity
Angular velocity tells us how fast (and in
what direction) something is spinning.
Δθ
r
The z subscript indicates the axis is in the
zdirection (and the rotation is therefore in the xy plane)
Counterclockwise is positive
Also have angular acceleration which describes how the
spinning rate changes
θ
Using Angular Variables
The angular position of a line on a disk of radius
6 cm is given by θ = 10 − 5t + 4t 2 rad.
Find the average angular speed between 1 and
3 s:
θ (1) = 10 − 5 + 4 = 9rad.
€
θ (3) = 10 −15 + 36 = 31rad.
θ (3) − θ (1) 31− 9
ω =
=
= 11rad /s
3 −1
2
Using Angular Variables
The angular position of a line on a disk of radius
2
6 cm is given by
θ = 10 − 5t + 4t rad.
Find the linear speed of a point on the rim at 2 s:
dθ
ω = ;ω (2)
€ = −5 + 8t = 11rad /s
dt
v = rω = .(06m)(11rad /s)
v(2) = .66m /s
Using Angular Variables
The angular position of a line on a disk of radius
2
6 cm is given by
θ = 10 − 5t + 4t rad.
Find the radial and tangential acceleration of a
point on the rim at 2 s:
dθ
= −5 +€8t
dt
2
dω d θ
2
α=
= 2 ;α (2) = 8rad /s
dt dt
2
atan = rα = .(06m)(8rad /s )
atan (2) = .48m /s2
€
2
2
atotal = arad + atan
2
v
(.66m /s)
arad = =
r
.06m
arad (2) = 7.26m /s2
2
Clicker question 3
Set frequency to BA
Big Ben and a little alarm clock (synchronized to an atomic
clock) both keep perfect time. Which minute hand has the
largest angular velocity?
A.  Big Ben
B.  little alarm clock
C.  same
D.  Impossible to tell
Clicker question 3
Set frequency to BA
Big Ben and a little alarm clock (synchronized to an atomic
clock) both keep perfect time. Which minute hand has the
largest angular velocity?
A.  Big Ben
B.  little alarm clock
C.  same
D.  Impossible to tell
Angular velocity is only a
measure of how quickly
the angle changes. Both
minute hands complete 1
revolution every hour.
1 rph = 2π rad/hr = 2π/3600 rad/s
x(cm) = (m1 x1 + m2 x + m3 x3 + m4 x4 ) / (mtotal)
Numbering the squares 1, 2, and 3 walking down, and #4 is the one on
the left, gives x(cm) = m*(-a/2)+m*(-a/2)+m*(-a/2)+m*(-3/2a)/ (4m) =
-3a/4.
Angular velocity & acceleration vectors
Note the subscript z on the angular velocity and
acceleration which indicates the axis of rotation
Also, note that ω and α can be positive or negative
This is like 1D motion. Is there an equivalent 3D motion?
Yes. The axis of rotation can change orientation; it is not
always along the z-axis (and therefore neither is ω).
Angular velocity and acceleration are vectors:
points perpendicular to the plane of rotation in the
direction given by the right hand rule (direction of your
thumb when fingers curl in direction of rotation).
Summary of angular kinematics
Angular displacement Δθ measured in radians
Δθ
Angular velocity:
Angular acceleration:
Angular velocity and acceleration are vectors:
points perpendicular to the plane of rotation in the
direction given by the right hand rule (direction of your
thumb when fingers curl in direction of rotation).
r
θ
Angular kinematics
The same equations which were derived for constant
linear (or translational) acceleration apply for constant
angular (or rotational) acceleration
Constant linear
acceleration only!
Constant angular
acceleration only!
Clicker question 4
Set frequency to BA
A flywheel with a mass of 120 kg, and a radius of 0.6 m, starting
at rest, has an angular acceleration of 0.1 rad/s2. How many
revolutions has the wheel undergone after 10 s?
What equation should the student use to obtain the answer?
A. 
B. 
C. 
D. 
E.  None of them will work
Clicker question 4
Set frequency to BA
A flywheel with a mass of 120 kg, and a radius of 0.6 m, starting
at rest, has an angular acceleration of 0.1 rad/s2. How many
revolutions has the wheel undergone after 10 s?
What equation should the student use to obtain the answer?
A. 
B. 
C. 
D. 
θ0=0, ω0z=0, αz=0.1 rad/s2 and we want θ
E.  None of them will work
Relationships to linear velocity
If we want the linear displacement or velocity of a point on a
rotating object, we need r and either θ or ω.
What is the speed of the tire rim if the radius
r s
is 0.35 m and the tire rotates at 20 rad/s?
θ
Radians measure distance around a unit circle
Therefore
(only for θ in radians!)
Time derivative of both sides (r is constant):
. Linear speed is radius times angular speed
Rim speed is
Relationships to linear acceleration
What about acceleration? If speed is
then
We know that centripetal (or radial) acceleration is
Using v = rω, this can be rewritten as
.
.
Total linear acceleration is composed of both tangential and radial
acceleration which are always perpendicular to each other.
Clicker question 5
Set frequency to BA
CD slows from an angular velocity of 4 rad/s
to a stop in 4 seconds with constant angular
acceleration. What is the magnitude of the
angular acceleration (α) of the CD?
A.  1 rad/s2
B.  4 rad/s2
C.  16 rad/s2
D.  1 cm/s2
E.  None of the above
5 cm
Clicker question 5
Set frequency to BA
CD slows from an angular velocity of 4 rad/s
to a stop in 4 seconds with constant angular
acceleration. What is the magnitude of the
angular acceleration (α) of the CD?
A.  1 rad/s2
B.  4 rad/s2
C.  16 rad/s2
D.  1 cm/s2
E.  None of the above
5 cm
Remember
and for
constant α we have
but want magnitude: 1 rad/s2.
Clicker question 6
Set frequency to BA
CD has an angular acceleration of -1 rad/s2 and an
angular velocity of 2 rad/s. What is the magnitude
5 cm
of the acceleration of the dust particle 5cm out?
A.  20 cm/s2
B.  5 cm/s2
C.  21 cm/s2
Total acceleration is composed of
D.  0.8 cm/s2
E.  None of the above radial and tangential acceleration.
Can now get net force by multiplying acceleration by mass